Which integral gives the arc length of the graph of $f(x)=\dfrac1{x^2}$ over the interval $[1, k]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_1^k \sqrt{1+\dfrac4{x^6}}~dx$ (Choice B) B $ \int_1^k \sqrt{1-\dfrac2{x^3}}~dx$ (Choice C) C $ \int_1^k \sqrt{1+\dfrac1{x^2}}~dx$ (Choice D) D $ \int_1^k \sqrt{1+\dfrac1{x^4}}~dx$
Solution: The arc length $L$ of the graph of the function $f$ over the interval $[a, b]$ is $ L = \int_a^b \sqrt{1+\big[f\,^\prime(x)\big]^2} dx$. First, calculate $f\,^\prime(x)$. $\begin{aligned} f(x) &= \dfrac1{x^2} \\\\ f\,^\prime(x) &= -\dfrac2{x^3} \end{aligned}$ Now apply the arc length formula on the interval $[1,k]$ and simplify the integral. $\begin{aligned} L&=\int_1^k \sqrt{1+\left(-\dfrac2{x^3}\right)^2}~dx \\\\ &=\int_1^k \sqrt{1+\dfrac4{x^6}}~dx \end{aligned}$